If there are any problems or mistakes, just correct them....
Equilibrium: 2a) 1.0 x 10^-4; b) 4.5 x 10^-3; c) 4500L/mol; d) 4.0 L/mol; e) 0.15 mol^-2; f) 0.4 mol^-2;
g) 1.02 x 10^5 L/mol; h) 50 L/mol; i) 0.116 mol/L; j) 4.94 x 10^-4 mol/L; k) 0.063M
3) 4.68 L/mol; 4) 0.024 mol/L; 5) 0.057M; 6) 305 M^-2; 7) 0.8 M; 8) Kconstant is 852; 9) a 1.005M^2
9b) 1.002M
Writing Equilibrium Expressions:
1) 0.37M; 2) 0.16 M^-2; 3) 20M^-0.5; 4) 0.133L/mol; 5) 421M^2; 6) K equilibrium is 59.9
7) 6.41 x 10^5 L/mol; 8) 0.0224 mol/L; 9) 1.29; 10) 0.37 M; 11) 139L/mol; 12) Kequilibrium average is 49.25 For #12 i got 41.2 Im not sure if maybe i made a calculation error though.-aditi
Equilibrium Calculations Using the Quadratic Equation (some may not be)
1) x = 3.009 x 10^-3 M
[N2] at eq.= [O2] eq. =0.054M
[NO] equ. = 0.006M
2) x = 1.003 x 10^-4 M
[N2] eq. = 0.057 M
[O2] eq. = 0.0539 M
[NO] = 0.0062 M
3) x = 0.0053 M
[N2]=[O2]eq.= 0.0947 M
[NO] eq. = 0.011 M
4) x = 0.399 M
[CO] eq. = 0.001 M
[Cl2]eq. = 0.301 M
[COCl2] eq. = 0.399 M
5) [H2]= [F2] = 3.16 x 10^-48 M
The decomposition does not occur to a great extent due to small Keq value.
6) x = 0.064 M
[PCl3] eq.= [Cl2] eq= 0.036 M
[PCl5] eq. = 0.064 M
7) Kconstant is 128.4
I could just be being a tard, but i know that a few other people keep getting 128 as an answer too for this question... Anyone able to guide? -Kat
You are right about answer being 128. I found my mistake...Thanks
Few More problems:**
1 a) forward reaction favoured since increasing the pressure decreases the volume and less moles are present in product side. Equilibrium shifts towards products.
b) Forward reaction favoured and equilibrium shifts towards products.
c) Reverse reaction favoured and equilibrium shifts towards reactants since increase in temp. causes enfothermic reaction to be favoured.
2a) Endothermic reaction is favoured and therefore decreasing [N2O4].
b) Forward reaction favoured and therefore increasing the [N2O4].
3a) Reverse reaction favoured as adding NH4Cl increases [NH4+] and therefore [NH4OH] goes up.
b) No change.
c) Forward reaction is favoured since adding HCl increases H+ concentration. These ions will neutralize OH-, which will decrease [ OH-]. Since [OH-] goes down, forward reaction favoured.
d) Reverse reaction favoured. Adding NaOH increases [OH-]. This causes the [NH4OH] to increase.
e) Forward reaction is favoured. Pure water decreases the concentration of substrate and products; therefore diluting them. This will decrease the value for Kequilibrium. In order to reach actual Keq. the concentration of products have to be greater than concentration of reactants. Since Keq. = ([NH4 +] [OH-]) / [NH4OH]
(It is possible for NH4OH to be aqueous yet unionized).
4) a precipitate of KCl will form since at saturation, Ksp is equal to Trial Ksp.
Equilibrium:
2a) 1.0 x 10^-4; b) 4.5 x 10^-3; c) 4500L/mol; d) 4.0 L/mol; e) 0.15 mol^-2; f) 0.4 mol^-2;
g) 1.02 x 10^5 L/mol; h) 50 L/mol; i) 0.116 mol/L; j) 4.94 x 10^-4 mol/L; k) 0.063M
3) 4.68 L/mol; 4) 0.024 mol/L; 5) 0.057M; 6) 305 M^-2; 7) 0.8 M; 8) Kconstant is 852; 9) a 1.005M^2
9b) 1.002M
Writing Equilibrium Expressions:
1) 0.37M; 2) 0.16 M^-2; 3) 20M^-0.5; 4) 0.133L/mol; 5) 421M^2; 6) K equilibrium is 59.9
7) 6.41 x 10^5 L/mol; 8) 0.0224 mol/L; 9) 1.29; 10) 0.37 M; 11) 139L/mol; 12) Kequilibrium average is 49.25
For #12 i got 41.2 Im not sure if maybe i made a calculation error though. -aditi
Equilibrium Calculations Using the Quadratic Equation (some may not be)
1) x = 3.009 x 10^-3 M
[N2] at eq.= [O2] eq. =0.054M
[NO] equ. = 0.006M
2) x = 1.003 x 10^-4 M
[N2] eq. = 0.057 M
[O2] eq. = 0.0539 M
[NO] = 0.0062 M
3) x = 0.0053 M
[N2]=[O2]eq.= 0.0947 M
[NO] eq. = 0.011 M
4) x = 0.399 M
[CO] eq. = 0.001 M
[Cl2]eq. = 0.301 M
[COCl2] eq. = 0.399 M
5) [H2]= [F2] = 3.16 x 10^-48 M
The decomposition does not occur to a great extent due to small Keq value.
6) x = 0.064 M
[PCl3] eq.= [Cl2] eq= 0.036 M
[PCl5] eq. = 0.064 M
7) Kconstant is 128.4
I could just be being a tard, but i know that a few other people keep getting 128 as an answer too for this question... Anyone able to guide?
-Kat
You are right about answer being 128. I found my mistake...Thanks
Few More problems:**
1 a) forward reaction favoured since increasing the pressure decreases the volume and less moles are present in product side. Equilibrium shifts towards products.
b) Forward reaction favoured and equilibrium shifts towards products.
c) Reverse reaction favoured and equilibrium shifts towards reactants since increase in temp. causes enfothermic reaction to be favoured.
2a) Endothermic reaction is favoured and therefore decreasing [N2O4].
b) Forward reaction favoured and therefore increasing the [N2O4].
3a) Reverse reaction favoured as adding NH4Cl increases [NH4+] and therefore [NH4OH] goes up.
b) No change.
c) Forward reaction is favoured since adding HCl increases H+ concentration. These ions will neutralize OH-, which will decrease [ OH-]. Since [OH-] goes down, forward reaction favoured.
d) Reverse reaction favoured. Adding NaOH increases [OH-]. This causes the [NH4OH] to increase.
e) Forward reaction is favoured. Pure water decreases the concentration of substrate and products; therefore diluting them. This will decrease the value for Kequilibrium. In order to reach actual Keq. the concentration of products have to be greater than concentration of reactants. Since Keq. = ([NH4 +] [OH-]) / [NH4OH]
(It is possible for NH4OH to be aqueous yet unionized).
4) a precipitate of KCl will form since at saturation, Ksp is equal to Trial Ksp.